(You may see the rest of the lecture notes here).

Part II: De Giorgi-Nash-Moser theory.

Throughout this lecture we make the following assumptions: we have a domain $\Omega$ that is bounded and has a smooth ($C^2$) boundary, and a symmetric matrix field $A(x)$ in $\Omega$ which is a measurable function in $x$ and such that for some positive constants $\lambda,\Lambda$ we have \begin{align} \lambda \text{I} \leq A(x) \leq \Lambda \text{I} \text{ for } a.e. x\in\Omega. \end{align}

Then, we will study the regularity of solutions to the Dirichlet problem \begin{align} \text{div}(A\nabla u) & = 0 \text{ in } \Omega,\\ \
u & = g \text{ on } \partial \Omega. \end{align}

We will be working with weak solutions $u \in H^1(\mathbb{R}^n)$, $u_{\mid \partial \Omega} = g$ (understood in the trace sense). The function $u$ will be said to be a weak solution if for any $\phi \in C^\infty_c(\Omega)$ we have \begin{align} -\int_{\Omega}A(x)\nabla u\cdot \nabla \phi \;dv = 0. \end{align}

The gist of De Giorgi-Nash-Moser theory is that a function $u$ having this property must be Holder continuous with exponent $\alpha$ for some $\alpha \in (0,1)$ determined entirely by $n$,$\lambda$, and $\Lambda$.

Subsolutions and supersolutions

Definition. A function $u \in H^1(\Omega)$ is said to be a (weak) subsolution if \begin{align} -\int_{\Omega}A(x)\nabla u\cdot \nabla \phi \;dv \geq 0, \end{align} for any non-negative $\phi \in C^\infty_c(\Omega)$. A (weak) supersolution corresponds to the opposite inequality, namely, \begin{align} -\int_{\Omega}A(x)\nabla u\cdot \nabla \phi \;dv \leq 0, \end{align} for any non-negative $\phi \in C^\infty_c(\Omega)$.

Exercise. Show that $u \in H^1(\Omega)$ with $u\mid_{\partial \Omega} = g$ is a weak solution if and only if it is both a weak subsolution and a weak supersolution.

Exercise. Show that the definition of weak subsolution (supersolution) is equivalent to the one we obtain if we instead ask the inequality hold for any a.e. non-negative function $\phi \in H^1_0(\Omega)$.

Proposition. Consider a non-decreasing function $h:\mathbb{R}\to\mathbb{R}$ of class $C^2$ such that $h(r) = O(|r|)$ as $|r|\to \infty$, and consider a function a solution $u\in H^1(\Omega)$. Define the function $v(x) = h(v(x))$. Then, if $ h''\geq 0$ the function $v$ will be a subsolution, while if $h''\leq 0$ the function $v$ will be a supersolution.

Proof. The proofs are entirely parallel, so we focus on the case $h''\geq 0$. Fix an arbitrary function $\phi \in C^\infty_c(\mathbb{R}^n)$ and observe that \begin{align} A(x)h'(u)\nabla u\cdot \nabla \phi = A(x)\nabla u\cdot (h'(u)\nabla \phi). \end{align} On the other hand, $h'(u)\nabla \phi = \nabla (h'(u)\phi) - \phi h''(u)\nabla u$, so \begin{align} -A h'(u)\nabla u\cdot \nabla \phi = -A\nabla u\cdot (\nabla (h'(u)\phi)) + \phi h''(u) (A\nabla u,\nabla u) \end{align} Since $h''(u)\geq 0$ a.e. we conclude that \begin{align} -A h'(u)\nabla u\cdot \nabla \phi \geq -A\nabla u\cdot (\nabla (h'(u)\phi)) \end{align} We integrate both sides of the equation. Observe that $h'(u)\phi$ belongs to $H^1\_0(\mathbb{R}^n)$ if $\phi \in C^\infty_c(\mathbb{R}^n)$, so from the fact that $u$ is a weak solution we have \begin{align} -\int_{\Omega} A h'(u)\nabla u\cdot \nabla \phi\;dx \geq -\int_{\Omega} A\nabla u\cdot (\nabla (h'(u)\phi))\;dx = 0 \end{align} and we conclude $u$ is a subsolution.

Exercise. Suppose $u_k$ is a sequence of functions in $H^1(\Omega)$ each of them a subsolution (resp. supersolution). Then, if $u_k$ converges weakly in $H^1$ to a function $v \in H^1(\Omega)$, the function $v$ will be a subsolution (supersolution) as well.

A few important examples where the last proposition comes handy.

1. Consider the function $h(r) = (r-\alpha)_+ := \max\{r-\alpha,0\}$ for some $\alpha \in \mathbb{R}$. This function is non-decreasing and weakly convex, and it can be approximated by a sequence of $C^2$ non-decreasing convex functions. Then, if $u \in H^1(\Omega)$ is a solution, the “cut-off above $\alpha$” $v(x) := (u(x)-\alpha)_+$ will be a subsolution.
2. If instead $h(r) = (r-\alpha)_- := \min\{r-\alpha,0\}$ then this function $h$ is weakly concave, so arguing as in the previous example we see that if $u$ is a solution then $v(x) = (u(x)-\alpha)_-$ will be a supersolution – and additioanlly, $\tilde v(x) = (\alpha-u(x))_+$ will be a subsolution.
3. Consider now the function $h(r) = \log(r)$, restricted to $r>0$. This function is non-decreasing and $h''\leq 0$. Then, if $u \in H^1(\Omega)$ is a solution which is $>0$ a.e., it follows that the function $v(x) := \log(u(x))$ is a supersolution whenever $u$ is a positive solution.

Sobolev and Caccioppoli

We are going to describe now the two inequalities with different homogeneities that will play here the role played by the minimiality of $E$ and the isoperimetric inequality in the previous lecture.

The first is known as Caccioppoli’s inequality, and also sometimes called the energy inequality.

Lemma (Caccioppoli's inequality). Suppose $u$ is a weak subsolution. Then, given $\eta \in C^2_c(\Omega)$, and $\alpha \in \mathbb{R}$, we have \begin{align} \lambda \int_{\Omega} |\nabla (u-\alpha)_+|^2\;dx \leq \Lambda \int_{\Omega} |\nabla \eta|^2(u-\alpha)_+^2\;dx. \end{align} If instead $u$ were a weak supersolution, the inequality would be \begin{align} \lambda \int_{\Omega} |\nabla (u-\alpha)_-|^2\;dx \leq \Lambda \int_{\Omega} |\nabla \eta|^2(u-\alpha)_-^2\;dx. \end{align}

Proof. We observe that $-u$ is a subsolution if $u$ is a supersolution and that \begin{align} (u-\alpha)_- = \min\{u-\alpha,0\} = -\max \{\alpha-u,0\} = -(\alpha-u)_+. \end{align} Therefore, that the first inequality holds for any subsolution implies the second inequality holds for any supersolution. We then the inequality for subsolutions. Fix $\eta \in C^\infty_c(\Omega)$, and take $\phi = \eta^2(u-\alpha)_+$ as a test function, then \begin{align} -\int_{\Omega}A\nabla u\cdot \nabla (\eta^2 (u-\alpha)_+)\;dx \geq 0. \end{align} The integrand vanishes a.e. in $\{u\leq \alpha\}$, and a.e. in this set we have $\nabla u = \nabla (u-\alpha)_+$, so this inequality is equivalent to \begin{align} \int_{\Omega}(A\nabla (u-\alpha_+)) \cdot \nabla (\eta^2 (u-\alpha)_+)\;dx \leq 0. \end{align} Why use $\eta^2$ and not $\eta$? Well, observe that \begin{align} \nabla (\eta^2 (u-\alpha)_+) = 2(u-\alpha)_+ \eta \nabla \eta + \eta^2 \nabla (u-\alpha)_+, \end{align} and so one can complete the square. This means that the inner product \begin{align} (A\nabla (u-\alpha_+)) \cdot \nabla (\eta^2 (u-\alpha)_+), \end{align} can be rewritten as, \begin{align} (A\nabla (\eta(u-\alpha)_+))\cdot \nabla (\eta(u-\alpha)_+) - (A\nabla \eta,\nabla \eta)(u-\alpha)_+^2. \end{align} Substituting this in the integral inequality we conclude that \begin{align} \int_{\Omega}(A\nabla (\eta(u-\alpha)_+)) \cdot \nabla (\eta (u-\alpha)_+)\;dx \leq \int_{\Omega}(A\nabla \eta,\nabla \eta)(u-\alpha)_+^2\;dx \end{align} Lastly, since $\lambda \text{I} \leq A(x) \leq \Lambda \text{I}$ a.e., it follows that \begin{align} \lambda \int_{\Omega}|\nabla (\eta(u-\alpha)_+)|^2\;dx \leq \Lambda \int_{\Omega}|\nabla u|^2(u-\alpha)_+^2\;dx \end{align}

Corollary. Suppose $u$ is a subsolution. There is a universal constant $C$ such that for any $\alpha \in \mathbb{R}$ and any. two concentric balls $B_{\rho}(x_0)\subset B_{\rho'}(x_0) \subset \Omega$ we have \begin{align} \int_{B_{\rho}(x_0)}|\nabla (u-\alpha)_+|^2\;dx \leq \frac{C}{(\rho'-\rho)^2}\int_{B_{\rho'}(x_0)}(u-\alpha)^2_+\;dx. \end{align} Likewise, if $u$ is a supersolution, then \begin{align} \int_{B_{\rho}(x_0)}|\nabla (u-\alpha)_-|^2\;dx \leq \frac{C}{(\rho'-\rho)^2}\int_{B_{\rho'}(x_0)}(u-\alpha)^2_-\;dx. \end{align}

In particular, if $u$ is a solution, the two sets of inequalities hold. De Giorgi emphasizes in his original presentation that these inequalities are the only information coming from the fact “$u$ is a weak solution to the PDE” that are used in proving the Holder continuity of $u$.

The other inequality we will need is the Sobolev embedding, an inequality that holds for any function in $H^1(\mathbb{R}^n)$, and as such plays a role parallel to the isoperimetric inequality in the example about minimal boundaries we discussed in the previous lecture.

Theorem (Sobolev inequality). If $u \in H^1(\mathbb{R}^n)$, then \begin{align} ||u||_{2^*} \leq C_n ||\nabla u||_2,\; \text{ where } 2^* := \frac{2n}{n-2}. \end{align}

We also have a “localized” version of Sobolev’s inequality.

Lemma (Sobolev inequality, localized) There is a dimensional constant $C_n$ such that given any $u\in H^1(\mathbb{R}^n)$ and any ball $B=B_r(x_0)$ we have \begin{align} ||u||_{L^{2^*}(B)}^2 \leq C_n\left ( ||\nabla u||_{L^2(B)}^2 + r^{-2}||u||_{L^2(B)}^2 \right ). \end{align}

De Giorgi’s in-measure-to-uniform estimate

We are now set to show a weak solution $u \in H^1(\Omega)$ must be bounded in every compact subset of $\Omega$. As we have been saying, the proof will follow the “two inequalities of different homogeneity” argument we discussed in the previous lecture in the context of minimal surfaces.

Exercise: Suppose $E_0,E_1,E_2,\ldots$ is an infinite sequence of positive real numbers and that there are positive numbers $b,\delta$ such that \begin{align} E_k \leq b^k E_{k-1}^{1+\delta} \text{ for } k=1,2,\ldots \end{align} Then, \begin{align} E_0 < 1 \Rightarrow \lim \limits_{k\to \infty}E_k = 0. \end{align}

Theorem (De Giorgi's $L^2 \to L^\infty$ estimate). There is a constant $C = C(n,\lambda,\Lambda)$ such that if $B_{2r}(x_0)\subset \Omega$, then any subsolution $u$ satisfies the estimate \begin{align} \sup \limits_{B_r(x_0)}u_+(x)^2 \leq \frac{C}{r^n}\int_{B_{2r}(x_0)}u(x)^2\;dx. \end{align} (Note: Strictly speaking, we should be using ess-sup on the left hand side)

Proof. We introduce the notation \begin{align} D(\alpha,\rho) = \left \{ x\in B_\rho(x_0) : u(x)>\alpha \right \} \end{align} The conclusion of the theorem is equivalent to the statement \begin{align} \int_{D(\alpha,r/2)}(u-\alpha)_+^2\;dx = 0 \text{ for } \alpha = \frac{C}{r^n}\int_{B_r(x_0)}u(x)^2\;dx. \end{align} We proceed to use the Caccioppoli and Sobolev inequalities to show how the $L^2$ norm of $(u-\alpha)_+$ over $B_\rho$ decays if $\alpha$ increases whiel $\rho$ decreases. First, we apply the localized Sobolev embedding to $\phi = (u-\alpha)_+$ in the ball $B_\rho(x_0)$ for some $\alpha,\rho,$ and $x_0$ such that $B_{\rho}(x_0)\subset \Omega$. We have \begin{align} ||(u-\alpha)_+||_{L^{2^*}(B_\rho(x_0))}^2 \leq C||\nabla (u-\alpha)_+||_{L^2(B_\rho(x_0))}^2+C\rho^{-2}||(u-\alpha)_+||^2_{L^2(B_{\rho}(x_0))}. \end{align} On the other hand, Holder's inequality says that for any $p>2$ we have \begin{align} \int_{B_\rho(x_0)}(u-\alpha)^2_+\;dx & = \int_{D(\alpha,\rho)}(u-\alpha)^2_+\;dx \\ & \leq |D(\alpha,\rho)|^{1-\frac{2}{p}}\left (\int_{B_\rho(x_0)}(u-\alpha)^p_+\;dx\right )^{\frac{2}{p}} \end{align} We combine this last inequality with $p=2^{*}$ and the localized Sobolev embedding, \begin{align} ||(u-\alpha)_+||_{L^2(B_\rho(x_0))}^2 \leq C |D(\alpha,\rho)|^{\frac{2}{n}} \left ( ||\nabla (u-\alpha)_+||_{L^2(B_\rho(x_0))}^2 + \rho^{-2}||(u-\alpha)_+||_{L^2(B_\rho(x_0))}^2 \right ) \end{align} On the other hand, if $\rho'>\rho$, the Corollary above says that \begin{align} \int_{D(\alpha,\rho)}|\nabla (u-\alpha)_+|^2\;dx \leq \frac{C}{(\rho'-\rho)^2} \int_{D(\alpha,\rho')}(u-\alpha)_+^2\;dx. \end{align} Then, we have \begin{align} \int_{D(\alpha,\rho)} (u-\alpha)_+^2\;dx \leq C\left (\frac{1}{(\rho'-\rho)^2}+\frac{1}{\rho^2} \right ) |D(\alpha,\rho)|^{\frac{2}{n}} \int_{D(\alpha,\rho')} (u-\alpha)_+^2\;dx \end{align} In conclusion, if $\rho'>\rho$ and $\alpha'<\alpha$, \begin{align} \int_{D(\alpha,\rho)} (u-\alpha)_+^2\;dx \leq C\left (\frac{1}{(\rho'-\rho)^2}+\frac{1}{\rho^2} \right )\frac{1}{(\alpha-\alpha')^{2\delta}}\left ( \int_{D(\alpha',\rho')} (u-\alpha')_+^2\;dx \right )^{1+\delta} \end{align} where $\delta = \frac{2}{n}$. We now set fix $M>0$ to be determined, and write for every $k\in\mathbb{N}$, \begin{align} \alpha_k = M\left (1-\frac{1}{2^k}\right),\;\;\rho_k = \rho\left ( \frac{1}{2}+\frac{1}{2^k} \right ). \end{align} To prove the theorem, we will now show that if $M$ is chosen large enough then \begin{align} E_k := \int_{D(\alpha_k,\rho_k)}(u-\alpha_k)_+^2\;dx \rightarrow 0 \text{ as } k \to \infty. \end{align} Observe that \begin{align} \rho_{k-1}-\rho_k = \rho 2^{-k},\; \alpha_k-\alpha_{k-1} = M2^{-k} \end{align} In particular, \begin{align} \frac{1}{(\rho_{k-1}-\rho_k)^2}+\frac{1}{\rho_k^2} \leq \frac{4^{k+1}}{\rho^2},\;\frac{1}{(\alpha_k-\alpha_{k-1})^2} \leq \frac{1}{M^2}2^k \end{align} We conclude that for any $k\geq 1$ (and with $b=8$) \begin{align} E_k \leq C b^k \frac{1}{\rho^2 M^{2\delta}} E_{k-1}^{1+\delta} \end{align} Now, observe: if we multiply both sides by $C \rho^{-2}M^{-2}$, \begin{align} \left (\frac{C}{\rho^2 M^{2}} E_k \right ) \leq b^k \left ( \frac{C}{\rho^2 M^2} \right ) \frac{C}{\rho^2 M^{2\delta}} E_{k-1}^{1+\delta} \end{align} This simplifies to (for some possibly larger $\tilde b$), \begin{align} \tilde E_k \leq \tilde b^k \tilde E_k^{1+\delta},\;\tilde E_k = \frac{C}{\rho^2 M^2}E_k. \end{align} Now, we choose $M$ large enough so that $\tilde E_0\leq 1$ and apply the previous exercise to conclude $\tilde E_k \to 0$. We see $M$ can be chosen to be any number strictly bigger than \begin{align} \frac{C}\rho^2E_0. \end{align} We conclude that for $\alpha = \left ( \frac{C}{\rho^2}E_0 \right )^{\frac{1}{2}}$ we have \begin{align} \int_{D(\alpha,r/2)}(u-\alpha)^2\;dx = 0, \end{align} which means that \begin{align} u(x) \leq \left ( \frac{C}{\rho^2}E_0 \right )^{\frac{1}{2}} \text{ a.e. in } B_{r/2}(x_0). \end{align}